pushout of topological Hausdorff spaces is not Hausdorff $A$, $X$, $Y$ are topological Hausdorff spaces, $f:A\to X$, $g:A\to Y$ continuous maps. I search an example where the pushout $Z$ of the morphisms $f$ and $g$ is not Hausdorff.
I thought if I take $X=Y=\mathbb{R}$ endowed with the standard topology, $A=\mathbb{Q}$ endowed with the subspace-topology of $\mathbb{R}$ I have to choose $f$ and $g$ such that $Z=\frac{X\coprod Y}{f(a)\sim g(a)\; \forall a\in A}$ endowed with the quotient topology is homeomorphic to $\frac{\mathbb{R}}{\mathbb{Q}}$. Or I tried to construct something with $X=Y=S^1$.. Is my first approach ok? I'm not sure how to choose the maps, one of them is the inclusion, and the other map I don't know. Or could you give me an example? (I'm interested in something with $X=Y=S^1$, but not important).
 A: These separation properties often fail to be preserved when something is not closed. When $X=Y=S^1$, you can take $A$ to be the open arc $A=\{(a_1,a_2)\in S^1\mid a_2>0\}$, the map $f:A\to X$ is the inclusion, and $g:A\to Y$ is the inclusion as well. So we basically just glue the two copies of $S^1$ along the arc $A$. The constructed space is not Hausdorff.
A: Here’s a way to make your first idea work, though the pushout isn’t $\Bbb R/\Bbb Q$.
Let $f:\Bbb Q\to\Bbb R:x\mapsto x$ and $g:\Bbb Q\to\Bbb R:x\mapsto -x$. The resulting pushout is the quotient of $\{0,1\}\times\Bbb R$ by the relation $\langle i,x\rangle\sim\langle j,y\rangle$ iff $\langle i,x\rangle=\langle j,y\rangle$, or $i\ne j$ and $x=-y\in\Bbb Q$. This space is not Hausdorff: if $\langle q_n:n\in\Bbb Q\rangle$ is a sequence of rationals converging to the irrational $\alpha$ in $\Bbb R$, then
$$\Big\langle[\langle 0,q_n\rangle]:n\in\Bbb N\Big\rangle$$
converges to both $[\langle 0,\alpha\rangle]$ and $[\langle 1,-\alpha\rangle]$ in this space.
Other more familiar examples are the line with two origins and the branching line.
